Set up, and evaluate the integral that represents the surface area of rotation of this cardioid about the y-axis.

Elix

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Dec 7, 2018
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I do not know if I set up the actual equation correctly so any tips would help! Thanks!

Cardioid:
\(\displaystyle r=1+sin(theta)\)

Here is my work, I split it into smaller segments to keep it organized:

image1.jpeg
 

MarkFL

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I don't see any errors in your work so far. Let's write (after some simplification):

\(\displaystyle A=2\sqrt{2}\pi\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos(\theta)(\sin(\theta)+1)^{\frac{3}{2}}\,d\theta\)

Now it appears the integration should be straightforward. :)
 

MarkFL

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To follow up, let's let:

\(\displaystyle u=\sin(\theta)+1\implies du=\cos(\theta)\,d\theta\)

And our definite integral becomes:

\(\displaystyle A=2\sqrt{2}\pi\int_0^2 u^{\frac{3}{2}}\,du=2\sqrt{2}\pi\left(\frac{2}{5}2^{\frac{5}{2}}\right)=\frac{32\pi}{5}\)
 

Elix

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Dec 7, 2018
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To follow up, let's let:

\(\displaystyle u=\sin(\theta)+1\implies du=\cos(\theta)\,d\theta\)

And our definite integral becomes:

\(\displaystyle A=2\sqrt{2}\pi\int_0^2 u^{\frac{3}{2}}\,du=2\sqrt{2}\pi\left(\frac{2}{5}2^{\frac{5}{2}}\right)=\frac{32\pi}{5}\)
Thanks a bunch! I struggled to connect how you got to that simplification at first but figured it out. I really appreciate your help with this and my previous question!
 
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